1.

If \(\rm (2^x-1)^2=4(2^x-2)\) , then the value of x is 1. \(\rm log_23\)2. \(\rm log_32\)3. \(\rm log_25\)4. \(\rm log_53\)

Answer» Correct Answer - Option 1 : \(\rm log_23\)

Concept:

\(\rm \frac{log a}{log b}=log_ba\)

 

Calculation:

Here, \(\rm (2^x-1)^2=4(2^x-2)\)

\(⇒ \rm 2^{2x}-(2)2^x+1=(4)2^x-8\)

\(⇒ \rm 2^{2x}-(6)2^x+9=0\)

Now, let 2x = y

∴\(\rm y^2-6y+9=0\)

\(⇒ \rm y^2-3y-3y+9=0\)

\(⇒ \rm y(y-3)-3(y-3)=0\)

(y - 3) (y - 3) = 0

⇒ y = 3

⇒ 2x = 3

x log 2 = log(3)

⇒ x = \(\rm \frac{log 3}{log2}\)

= \(\rm log_23\)

Hence, option (1) is correct.


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